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anonymous

  • one year ago

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  1. campbell_st
    • one year ago
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    well an easy method is to graph the curve... and look where a tangent to the curve would have a negative slope

  2. campbell_st
    • one year ago
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    the other is to use calculus to sketch the curve

  3. ybarrap
    • one year ago
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    You are looking for where f'(x) < 0 Take the first derivative and see where this occurs

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  4. ybarrap
    • one year ago
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    Here is f'(x) $$ f'(x)=(x-6) (x-2) $$ When is this negative?

  5. anonymous
    • one year ago
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    At 6 and 2?

  6. ybarrap
    • one year ago
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    BETWEEN 6 and 2, right? |dw:1436130851705:dw|

  7. ybarrap
    • one year ago
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    |dw:1436130906919:dw|

  8. ybarrap
    • one year ago
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    $$ 2\lt x\lt 6 $$ Note that it does NOT include x=6 and x=2, because at these points f'(x)=0 (i.e. they are min and max respectively and not changin)

  9. anonymous
    • one year ago
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    So it would be (2,6) only?

  10. ybarrap
    • one year ago
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    Yes! At all the other points, slope is positive or zero. Remember that slope is what f'(x) means: |dw:1436131100676:dw|

  11. ybarrap
    • one year ago
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    |dw:1436131184837:dw|

  12. ybarrap
    • one year ago
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    I've been forgetting to put my " ' " in my picture for f'(x) (derivative)

  13. anonymous
    • one year ago
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    Alright thank you so much!!

  14. ybarrap
    • one year ago
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    You're welcome!!

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